operator-precedence

There's been a change in Swift 3 for Xcode 8 beta 6 and now I'm not able to declare my operator for power as I did before: infix operator ^^ { } public func ^^ (radix: Double, power: Double) -> Double { return pow((radix), (power)) } I've read a bit about it and there's a new change been introduced in Xcode 8 beta 6 From this I'm guessing I have to declare a precedence group and use it for my operator like this: precedencegroup ExponentiativePrecedence {} infix operator ^^: ExponentiativePrecedence public func ^^ (radix: Double, power: Double) -> Double { return pow((radix), (power)) } Am I

Defined this way, we can do neither ++x++ nor ++x-- . But on the other hand, both (++x)++ and (++x)-- are useful expressions: (++x)++ increments x by two and returns the value "in the middle", while (++x)-- is essentially equivalent to x+1 but completely avoids having to call operator+ , which can be quite useful sometimes. So why is the precedence not defined to have ++x++ automatically expand to (++x)++ rather than ++(x++) ? Is there some hidden meaning to the latter which I don't understand, or is it just to keep the precedence a simple list with all prefix operators making up one single

EDIT: Following Mike Seymour's comment, I replaced operator std::string () const; with operator char * () const; and changed the implementation accordingly. This allows implicit casting, but, for some reason, the unsigned long int operator has precedence over the char * operator, which just does not feel right... Also, I don't want to expose nasty C stuff like char * outside the class, when I have std::string. I have a hunch that my CustomizedInt class needs to inherit from some stuff in order to support the feature that I desire. Could anybody please elaborate Mike's comment regarding std:

Say I have this code: unsigned int func1(); unsigned int func2(); unsigned int func3(); unsigned int x = func1() | func2() | func3(); Does C++ guarantee that func1() will be called first, then func2(), and then func3()? Or is the compiler allowed to call the functions in any order it feels like? Also, is the compiler allowed to implement a short-circuit optimization here if it wants to? (e.g. if func1() returned ~0, could the compiler decide not to bother calling func2() or func3(), because it knows their return values can't possibly affect the value assigned to x?) No, there is no guarantee

I have executed the following code in Code::Blocks 10.05 on Windows 7. int a=0,b=0,c; c=a++&&b++; printf("\na=%d\nb=%d\nc=%d\n\n",a,b,c); The output I obtained is given below, a=1 b=0 c=0 This makes perfect sense because of short circuit evaluation. The expression a++ is post increment and 0 is returned to the logical and ( && ). Hence the part b++ is not evaluated since both 0 && 0 and 0 && 1 evaluates to 0 . But here arises my doubt. The precedence value of operators clearly states that ++ is having higher precedence over && . So my understanding was like this, both a++ and b++ are evaluated

Can some one tell why my php line break not working ( echoing ) ? I know i can write the code in a different way to make the line break work, but i want to know the reason behind this ? <?php $var1 = 3; echo "Addition = " . $var1 += 3 . "<br>"; echo "Subtraction = " . $var1 -= 3 . "<br>"; echo "Multiplication = " . $var1 *= 3 . "<br>"; echo "Division = " . $var1 /= 3 . "<br>"; ?> You can use commas, echo "Addition = " . $var1 += 3 , "<br>"; echo "Subtraction = " . $var1 -= 3 ,"<br>"; echo "Addition = " . $var1 *= 3 , "<br>"; echo "Addition = " . $var1 /= 3 ,"<br>"; Or wrap it in brackets: echo